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Identity:
${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$
Example:
Expand $$27x^3 + 8y^3 + z^3 - 18xyz$$.

Solution:

Let us write the expression of $$27x^3 + 8y^3 + z^3 - 18xyz$$ using the identity ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$.

${27x}^{3}+{8y}^{3}+{z}^{3}-12\mathit{xyz}={\left(3x\right)}^{3}+{\left(2y\right)}^{3}+{\left(z\right)}^{3}-3\left(3x\right)\left(2y\right)\left(z\right)$

$$=$$ $\left(3x+2y+z\right)\left({\mathit{9x}}^{2}+{\mathit{4y}}^{2}+{z}^{2}-\left(\mathit{3x}\right)\left(\mathit{2y}\right)-\left(\mathit{2y}\right)\left(z\right)-\left(3x\right)\left(z\right)\right)$

$$=$$ $\left(3x+2y+z\right)\left({\mathit{3x}}^{2}+{\mathit{2y}}^{2}+{z}^{2}-6\mathit{xy}-2\mathit{yz}-3\mathit{xz}\right)$
Important!
If ${a}^{3}+{b}^{3}+{c}^{3}=0$ then the identity ${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(a+b+c\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$ is rewritten as follows:

${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(0\right)\left({a}^{2}+{b}^{2}+{c}^{2}-\mathit{ab}-\mathit{bc}-\mathit{ac}\right)$

${a}^{3}+{b}^{3}+{c}^{3}-3\mathit{abc}=\left(0\right)$

${a}^{3}+{b}^{3}+{c}^{3}=3\mathit{abc}$
Example:
Evaluate $$8^3 - 5^3 - 3^3$$.

Solution:

By the identity if ${a}^{3}+{b}^{3}+{c}^{3}=0$, then ${a}^{3}+{b}^{3}+{c}^{3}=3\mathit{abc}$.

On comparing $$8^3 - 5^3 - 3^3$$ with ${a}^{3}+{b}^{3}+{c}^{3}=0$ we have $$a$$ $$=$$ $$8$$, $$b$$ $$=$$ $$5$$ and $$c$$ $$=$$ $$3$$.

Here $$a+b+c$$ $$=$$ $$8-5-3$$ $$=$$ $$0$$

Therefore, $$8^3 - 5^3 - 3^3$$ $$=$$ $$3 \times 8 \times 5 \times 3$$ $$=$$ $$360$$